2019届高考数学总复习模块三数列限时集训（十一）数列求和及数列的简单应用理.docx

限时集训（十一）数列求和及数列的简单应用基础过关1.已知各项均为正数且递减的等比数列an满足a3,32a4,2a5成等差数列,其前5项和S5=31.(1)求数列an的通项公式;(2)若等差数列bn满足b1=a4-1,b2=a3-1,求数列abn的前n项和Tn.2.已知数列an满足a1=a3,an+1-an2=32n+1,设bn=2nan.(1)求数列bn的通项公式;(2)求数列an的前n项和Sn.3.已知数列an为正项数列,a1=4,且对任意nN*,an+12-2an2=anan+1恒成立.(1)求数列an的通项公式;(2)若数列bn满足bn=1nlog2an,Tn为数列bn的前n项和,求证:Tn1.4.已知an是等比数列,数列bn满足b1=-2,b2=5,且a1b1+a2b2+anbn=2+(2n-3)4n.(1)求an的通项公式和前n项和Sn;(2)求bn的通项公式.能力提升5.已知数列an满足an+1+1=an+1an+2,an-1且a1=1.(1)证明数列1an+1是等差数列,并求出数列an的通项公式;(2)令bn=2nan+1,求数列bn的前n项和Sn.6.已知数列an,bn,其中a1=3,b1=-1,且满足an=12(3an-1-bn-1),bn=-12(an-1-3bn-1),nN*,n2.(1)求证:数列an-bn为等比数列;(2)求数列2nanan+1的前n项和Sn.限时集训(十一) 基础过关1.解:(1)设an的公比为q(0q0,an+1=2an,数列an是首项为4,公比为2的等比数列,an=2n+1.(2)证明:bn=1nlog2an=1n(n+1)=1n-1n+1,Tn=b1+b2+bn=11-12+12-13+1n-1n+1=1-1n+11.4.解:(1)a1b1+a2b2+anbn=2+(2n-3)4n,a1b1=2-4=-2,a1b1+a2b2=2+(4-3)42=18,则a2b2=20,又b1=-2,b2=5,a1=1,a2=4,an是等比数列,a2a1=4,an的通项公式为an=4n-1,an的前n项和Sn=1-4n1-4=4n-13.(2)由an=4n-1及a1b1+a2b2+anbn=2+(2n-3)4n,得b1+4b2+4n-1bn=2+(2n-3)4n,当n2时,b1+4b2+4n-2bn-1=2+(2n-5)4n-1,-得4n-1bn=2+(2n-3)4n-2-(2n-5)4n-1=(6n-7)4n-1,bn=6n-7.又当n=1时,b1=-2,不满足上式,bn的通项公式为bn=-2,n=1,6n-7,n2. 能力提升5.解:(1)an+1+1=an+1an+2,an-1且a1=1,1an+1+1=an+2an+1,即1an+1+1=(an+1)+1an+1,1an+1+1-1an+1=1,数列1an+1是等差数列.1a1+1=12,1an+1=12+(n-1)1,1an+1=2n-12,an=3-2n2n-1.(2)由(1)知bn=(2n-1)2n-1,则Sn=120+321+522+(2n-1)2n-1,2Sn=121+322+(2n-3)2n-1+(2n-1)2n,-Sn=1+22+222+22n-1-(2n-1)2n=1+22(1-2n-1)1-2-(2n-1)2n,Sn=-1+22-2n+1+(2n-1)2n=3-2n+1+(2n-1)2n=(2n-3)2n+3.6.解:(1)证明:an-bn=12(3an-1-bn-1)-12(an-1-3bn-1)=2(an-1-bn-1),又a1-b1=3-(-1)=4,所以an-bn是首项为4,公比为2的等比数列.(2)由(1)知,an-bn=2n+1.因为an+bn=12(3an-1-bn-1)+-12(an-1-3bn-1)=an-1+bn-1,a1+b1=3+(-1)=2,所以an+bn为常数列且an+bn=2,联立得an=2n+1,故2nanan+1=2n(2n+1)(2n+